Programming Forums > Script Programming > Python

Chuckiferd · May 3rd, 2008, 2:29 PM

Hello I am trying to create a program that involves writing a whole bunch of lists with information of its specific topic then grouping them together. Normally I would use a hash but they don't really let you create a whole bunch of variables that can be withdrawn later.

So my problem is when I nest them together, based on a random factor I want to delete so many of them from the back, but nothing I write seems to work

I can see doing all this with a whole bunch of if: loops but I know there has to be some easier way

here is my program(at least the part that pertains to it)
(PS. all the importing and preperations are done I simply didn't print it)

a=randint(6,11)
aa=['name', 'att', 'def', 1]
ab=['name', 'att', 'def', 1]
ac=['name', 'att', 'def', 1]
ad=['name', 'att', 'def', 1]
ae=['name', 'att', 'def', 1]
af=['name', 'att', 'def', 1]
ag=['name', 'att', 'def', 1]
ah=['name', 'att', 'def', 1]
ai=['name', 'att', 'def', 1]
aj=['name', 'att', 'def', 1]
abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj]
del abspirlist[a:]
shuffle(abspirlist)

Sane · May 3rd, 2008, 2:45 PM

Are you trying to turn:

abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj]

Into a list terminated at a random value, like:

abspirlist=[aa,ab,ac,ad,ae,af]

Why not do:

(Toggle Plain Text)

abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj] 
abspirlist = abspirlist[:a]

Where a is your random integer. Or do you want to keep the first dimension in tact, and pop elements from each array in the second dimension? In that case, just use a for loop, with the same code as above used on each nested list individually.

Chuckiferd · May 3rd, 2008, 4:27 PM

whoa, why is it suddenly working, I must of been doing something wrong. Weird I could of swore that, that was the first thing I tried.

Thanks Sane

PS. Is there a way to automaticaly count the number of contents, (or nestled lists) in a list, if not I am going to have to make a mod because it would make everything so much easier

Sane · May 3rd, 2008, 4:43 PM

So if you had:

(Toggle Plain Text)

abspirlist = [['name', 'att', 'def', 1], 
              ['name', 'att', 'def', 1], 
              ['name', 'att', 'def', 1]]

You want the number 12, for the total number of nested elements?

You can do that with a little fancy work, by generating a list of lengths (in this case: [4, 4, 4]).

Then feeding it to the sum function (sum([4,4,4]) evaluates 12).

Combining these two ideas gives us:

(Toggle Plain Text)

sum((len(x) for x in abspirlist))

Chuckiferd · May 3rd, 2008, 5:24 PM

cool then I can divide that by 4 to figure out how many nested lists there are.

You know ever since last night everything seems to be going wrong, I am adding the code to make each list unique and different. Look at whats happening

(Toggle Plain Text)

namelist=['Fred', 'Joe', 'Edward', 'Eric', 'John', 'Ryan', 'Chuck', 'Henry', 'Jean', 'Robert', 'Guido', 'Alan', 'Nicolas', 'Bill']
def whrstat():
  dice=randint(0, 13)
  whrname=namelist[dice]
  att=randint(1, 5)
  deff=randint(1, 5)

whrstat()
att                               x
deff                             y
aa=[whrname, att, deff, 1]
whrstat()                      
att                                x
deff                               y; ect. ect. ect.
ab=[whrname, att, deff, 1]
whrstat()
att
def
ac=[whrname, att, deff, 1]
whrstat()
att
def
ad=[whrname, att, deff, 1]
whrstat()
att
def
ae=[whrname, att, deff, 1]
whrstat()
att
def
af=[whrname, att, deff, 1]
whrstat()
att
def
ag=[whrname, att, deff, 1]
whrstat()
att
def
ah=[whrname, att, deff, 1]
whrstat()
att
def
ai=[whrname, att, deff, 1]
whrstat()
att
def
aj=[whrname, att, deff, 1]
abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj] 
dice=randint(0,4)
dice
del abspirlist[:dice]
shuffle(abspirlist)

Lets just say x=4 and y=3

(Toggle Plain Text)

>>>print abspirlist

[Guido, 4, 3, 1]

Sane · May 3rd, 2008, 5:33 PM

Wait, what? Hold on a tick. Divide by 4?

If you just want the number of lists, then use the len function...

(Toggle Plain Text)

len(abspirlist)

That will give you 3 in my example.

And in the code you posted, don't do:

(Toggle Plain Text)

del abspirlist[:dice]

Reassign your array in the same manner I showed you in my second post.

(Toggle Plain Text)

x = x[a:]   # Will keep all elements from the a'th element to the end
x = x[:a]   # Will keep all elements until the a'th element

Chuckiferd · May 3rd, 2008, 6:46 PM

got it that was actually old un-updated code, I fixed all that, but do you get why the number generator is not working

Sane · May 3rd, 2008, 9:37 PM

It's not that it's not working. What appears to be a lack of randomness is actually a side-effect of not creating new variables.

When you create variables inside a function, they need to be returned back to where the function was called if you want to access them.

(Toggle Plain Text)

from random import randint

namelist=['Fred', 'Joe', 'Edward', 'Eric', 'John', 'Ryan', 'Chuck', 'Henry', 'Jean', 'Robert', 'Guido', 'Alan', 'Nicolas', 'Bill']

def whrstat():
  dice=randint(0, 13)
  whrname=namelist[dice]
  att=randint(1, 5)
  deff=randint(1, 5)
  return [whrname, att, deff, 1]

aa = whrstat()
print aa

ab = whrstat()
print ab

(Toggle Plain Text)

['John', 4, 2, 1]
['Henry', 1, 2, 1]

What that will do is create the variables inside the function, return them as a list, and then assign them to whatever's calling the function.

If the explanation for this behaviour does not seem obvious to you, I suggest that you do a little review on how to use functions and return.

Chuckiferd · May 3rd, 2008, 10:40 PM

ok I see, I always get so caught up in one method that I never think of changing it.

May 3rd, 2008, 2:29 PM	#1
Chuckiferd Programmer Join Date: Nov 2007 Posts: 61 Rep Power: 1	Are nested lists indestructable? Hello I am trying to create a program that involves writing a whole bunch of lists with information of its specific topic then grouping them together. Normally I would use a hash but they don't really let you create a whole bunch of variables that can be withdrawn later. So my problem is when I nest them together, based on a random factor I want to delete so many of them from the back, but nothing I write seems to work I can see doing all this with a whole bunch of if: loops but I know there has to be some easier way here is my program(at least the part that pertains to it) (PS. all the importing and preperations are done I simply didn't print it) a=randint(6,11) aa=['name', 'att', 'def', 1] ab=['name', 'att', 'def', 1] ac=['name', 'att', 'def', 1] ad=['name', 'att', 'def', 1] ae=['name', 'att', 'def', 1] af=['name', 'att', 'def', 1] ag=['name', 'att', 'def', 1] ah=['name', 'att', 'def', 1] ai=['name', 'att', 'def', 1] aj=['name', 'att', 'def', 1] abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj] del abspirlist[a:] shuffle(abspirlist) __________________ There are 10 kinds of people in this world, those who can read binary and those who can't.

May 3rd, 2008, 2:45 PM	#2
Sane Programming Guru Join Date: Apr 2005 Posts: 1,722 Rep Power: 5	Re: Are nested lists indestructable? Are you trying to turn: `abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj]` Into a list terminated at a random value, like: `abspirlist=[aa,ab,ac,ad,ae,af]` Why not do: (Toggle Plain Text) abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj] abspirlist = abspirlist[:a] abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj] abspirlist = abspirlist[:a] Where a is your random integer. Or do you want to keep the first dimension in tact, and pop elements from each array in the second dimension? In that case, just use a for loop, with the same code as above used on each nested list individually. __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

May 3rd, 2008, 4:27 PM	#3
Chuckiferd Programmer Join Date: Nov 2007 Posts: 61 Rep Power: 1	Re: Are nested lists indestructable? whoa, why is it suddenly working, I must of been doing something wrong. Weird I could of swore that, that was the first thing I tried. Thanks Sane PS. Is there a way to automaticaly count the number of contents, (or nestled lists) in a list, if not I am going to have to make a mod because it would make everything so much easier __________________ There are 10 kinds of people in this world, those who can read binary and those who can't.

May 3rd, 2008, 4:43 PM	#4
Sane Programming Guru Join Date: Apr 2005 Posts: 1,722 Rep Power: 5	Re: Are nested lists indestructable? So if you had: (Toggle Plain Text) abspirlist = [['name', 'att', 'def', 1], ['name', 'att', 'def', 1], ['name', 'att', 'def', 1]] abspirlist = [['name', 'att', 'def', 1], ['name', 'att', 'def', 1], ['name', 'att', 'def', 1]] You want the number 12, for the total number of nested elements? You can do that with a little fancy work, by generating a list of lengths (in this case: [4, 4, 4]). Then feeding it to the sum function (sum([4,4,4]) evaluates 12). Combining these two ideas gives us: (Toggle Plain Text) sum((len(x) for x in abspirlist)) sum((len(x) for x in abspirlist)) __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

May 3rd, 2008, 5:24 PM	#5
Chuckiferd Programmer Join Date: Nov 2007 Posts: 61 Rep Power: 1	Re: Are nested lists indestructable? cool then I can divide that by 4 to figure out how many nested lists there are. You know ever since last night everything seems to be going wrong, I am adding the code to make each list unique and different. Look at whats happening (Toggle Plain Text) namelist=['Fred', 'Joe', 'Edward', 'Eric', 'John', 'Ryan', 'Chuck', 'Henry', 'Jean', 'Robert', 'Guido', 'Alan', 'Nicolas', 'Bill'] def whrstat(): dice=randint(0, 13) whrname=namelist[dice] att=randint(1, 5) deff=randint(1, 5) whrstat() att x deff y aa=[whrname, att, deff, 1] whrstat() att x deff y; ect. ect. ect. ab=[whrname, att, deff, 1] whrstat() att def ac=[whrname, att, deff, 1] whrstat() att def ad=[whrname, att, deff, 1] whrstat() att def ae=[whrname, att, deff, 1] whrstat() att def af=[whrname, att, deff, 1] whrstat() att def ag=[whrname, att, deff, 1] whrstat() att def ah=[whrname, att, deff, 1] whrstat() att def ai=[whrname, att, deff, 1] whrstat() att def aj=[whrname, att, deff, 1] abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj] dice=randint(0,4) dice del abspirlist[:dice] shuffle(abspirlist) namelist=['Fred', 'Joe', 'Edward', 'Eric', 'John', 'Ryan', 'Chuck', 'Henry', 'Jean', 'Robert', 'Guido', 'Alan', 'Nicolas', 'Bill'] def whrstat(): dice=randint(0, 13) whrname=namelist[dice] att=randint(1, 5) deff=randint(1, 5) whrstat() att x deff y aa=[whrname, att, deff, 1] whrstat() att x deff y; ect. ect. ect. ab=[whrname, att, deff, 1] whrstat() att def ac=[whrname, att, deff, 1] whrstat() att def ad=[whrname, att, deff, 1] whrstat() att def ae=[whrname, att, deff, 1] whrstat() att def af=[whrname, att, deff, 1] whrstat() att def ag=[whrname, att, deff, 1] whrstat() att def ah=[whrname, att, deff, 1] whrstat() att def ai=[whrname, att, deff, 1] whrstat() att def aj=[whrname, att, deff, 1] abspirlist=[aa,ab,ac,ad,ae,af,ag,ah,ai,aj] dice=randint(0,4) dice del abspirlist[:dice] shuffle(abspirlist) Lets just say x=4 and y=3 (Toggle Plain Text) >>>print abspirlist >>>print abspirlist [Guido, 4, 3, 1] __________________ There are 10 kinds of people in this world, those who can read binary and those who can't.

May 3rd, 2008, 5:33 PM	#6
Sane Programming Guru Join Date: Apr 2005 Posts: 1,722 Rep Power: 5	Re: Are nested lists indestructable? Wait, what? Hold on a tick. Divide by 4? If you just want the number of lists, then use the len function... (Toggle Plain Text) len(abspirlist) len(abspirlist) That will give you 3 in my example. And in the code you posted, don't do: (Toggle Plain Text) del abspirlist[:dice] del abspirlist[:dice] Reassign your array in the same manner I showed you in my second post. (Toggle Plain Text) x = x[a:] # Will keep all elements from the a'th element to the end x = x[:a] # Will keep all elements until the a'th element x = x[a:] # Will keep all elements from the a'th element to the end x = x[:a] # Will keep all elements until the a'th element __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

May 3rd, 2008, 6:46 PM	#7
Chuckiferd Programmer Join Date: Nov 2007 Posts: 61 Rep Power: 1	Re: Are nested lists indestructable? got it that was actually old un-updated code, I fixed all that, but do you get why the number generator is not working __________________ There are 10 kinds of people in this world, those who can read binary and those who can't.

May 3rd, 2008, 9:37 PM	#8
Sane Programming Guru Join Date: Apr 2005 Posts: 1,722 Rep Power: 5	Re: Are nested lists indestructable? It's not that it's not working. What appears to be a lack of randomness is actually a side-effect of not creating new variables. When you create variables inside a function, they need to be returned back to where the function was called if you want to access them. (Toggle Plain Text) from random import randint namelist=['Fred', 'Joe', 'Edward', 'Eric', 'John', 'Ryan', 'Chuck', 'Henry', 'Jean', 'Robert', 'Guido', 'Alan', 'Nicolas', 'Bill'] def whrstat(): dice=randint(0, 13) whrname=namelist[dice] att=randint(1, 5) deff=randint(1, 5) return [whrname, att, deff, 1] aa = whrstat() print aa ab = whrstat() print ab from random import randint namelist=['Fred', 'Joe', 'Edward', 'Eric', 'John', 'Ryan', 'Chuck', 'Henry', 'Jean', 'Robert', 'Guido', 'Alan', 'Nicolas', 'Bill'] def whrstat(): dice=randint(0, 13) whrname=namelist[dice] att=randint(1, 5) deff=randint(1, 5) return [whrname, att, deff, 1] aa = whrstat() print aa ab = whrstat() print ab (Toggle Plain Text) ['John', 4, 2, 1] ['Henry', 1, 2, 1] ['John', 4, 2, 1] ['Henry', 1, 2, 1] What that will do is create the variables inside the function, return them as a list, and then assign them to whatever's calling the function. If the explanation for this behaviour does not seem obvious to you, I suggest that you do a little review on how to use functions and return. __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

May 3rd, 2008, 10:40 PM	#9
Chuckiferd Programmer Join Date: Nov 2007 Posts: 61 Rep Power: 1	Re: Are nested lists indestructable? ok I see, I always get so caught up in one method that I never think of changing it. __________________ There are 10 kinds of people in this world, those who can read binary and those who can't.

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Are nested lists indestructable?